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We have 9 terms, middle term is 40^2. Express all other terms as 40-x:\(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\)\(=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+2)^2+(40+3)^2+(40+4)^2\). Now, when you expand these expressions applying \((x-y)^2=x^2-2x+y^2\) and \((x+y)^2=x^2+2x+y^2\) you'll see that \(-2xy\) and \(2xy\) cancel out and we'll get:

Approach #2:The sum of the squares of the first n positive integers is given by: \(1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}\) (note that it's highly unlikely that you'll need it on the real GMAT test). For example the sum of the first 3 positive integers: \(1^2+2^2+3^3=\frac{3(3+1)(2*3+1)}{6}=14\).

Now, we can calculate the sum of the squares of the first 44 positive integers and subtract from it the sum of the squares of the first 35 positive integers to get the answer: \(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\frac{44(44+1)(2*44+1)}{6}-\frac{35(35+1)(2*35+1)}{6}=14,460\).

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:\(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\)\(=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+1)^2+(40+3)^2+(40+4)^2\). Now, when you expand these expressions applying \((x-y)^2=x^2-2x+y^2\) and \((x+y)^2=x^2+2x+y^2\) you'll see that \(-2xy\) and \(2xy\) cancel out and we'll get:

We have 9 terms, middle term is 40^2. Express all other terms as 40-x:\(36^2+37^2+38^2+39^2+40^2+41^2+42^2+43^2+44^2=\)\(=(40-4)^2+(40-3)^2+(40-2)^2+(40-1)^2+40^2+(40+1)^2+(40+1)^2+(40+3)^2+(40+4)^2\). Now, when you expand these expressions applying \((x-y)^2=x^2-2x+y^2\) and \((x+y)^2=x^2+2x+y^2\) you'll see that \(-2xy\) and \(2xy\) cancel out and we'll get:

hm, before looking to ur solution, I solved the q. in this way-let a =36^2then we have a+(a+1)+(a+2) (a+3) (a+4) (a+5) (a+6) (a+7) (a+8)

I feel that it is just an arithmetic progression with mean=median

so the sum of these numbers are 9*(a+4)=9*40^2=14400

later I saw ur solution -=9*40^2+2*(4^2+3^2+2^2+1^2)=14,400+60=14,460we seem to be in the same way, but using my method how to come to +2*(4^2+3^2+2^2+1^2)?

The problem is that 36^2, 37^2, 38^2, 39^2, 40^2, 41^2, 42^2, 43^2, and 44^2 DOES NOT form an AP. You assumed that \(a+1=36^2+1=37^2\), \(a+2=36^2+2=38^2\), ... but that's not correct: \(37^2\neq{36^2+1}\), \(38^2\neq{36^2+2}\), ... .

It is not an arithmetic progression, so if you assume that it is, you will get the wrong answer. An arithmetic progression is 'equally spaced'. If you just look at the smallest perfect squares, you can see that they are not equally spaced: 1, 4, 9, 16, 25... In fact the spacing gets larger the further you get into this list.

You can use the spacing of perfect squares to answer this question. From the difference of squares, we have that

41^2 - 40^2 = (41 + 40)(41 - 40) = 81

So 41^2 = 40^2 + 81. Similarly, 42^2 = 41^2 + 83, and 39^2 = 40^2 - 79, and so on. Listing all of the values we need to sum:

It is not an arithmetic progression, so if you assume that it is, you will get the wrong answer. An arithmetic progression is 'equally spaced'. If you just look at the smallest perfect squares, you can see that they are not equally spaced: 1, 4, 9, 16, 25... In fact the spacing gets larger the further you get into this list.

You can use the spacing of perfect squares to answer this question. From the difference of squares, we have that

41^2 - 40^2 = (41 + 40)(41 - 40) = 81

So 41^2 = 40^2 + 81. Similarly, 42^2 = 41^2 + 83, and 39^2 = 40^2 - 79, and so on. Listing all of the values we need to sum:

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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